Proving identities, again.

**Furyan** · December 24th 2011, 08:46 AM

I’m sorry to ask for help again

, but I can’t prove the following identity:

$\dfrac{\tan^3\theta}{1 + \tan^2\theta} + \dfrac{\cot^3\theta}{1 + \cot^2\theta} \equiv \dfrac{1 -2\sin^2\theta\cos^2\theta}{\sin\theta\cos\theta}$

I have tried but I’m afraid I simply can’t do it.

I keep getting:

$LHS \equiv \dfrac{\sin^4\theta + cos^4\theta}{\sin\theta\cos\theta}$

And I can’t see how to equate the numerators. I think I've probably made a mistake.

A hint would be very much appreciated.

Thank you.

**Siron** · December 24th 2011, 09:03 AM

Note that:
$\sin^4(\theta)+\cos^4(\theta)=\sin^4(\theta)+[1-\sin^2(\theta)]^2=\ldots$

**Furyan** · December 24th 2011, 09:20 AM

Thank you Siron

Originally Posted by Siron

Note that:
$\sin^4(\theta)+\cos^4(\theta)=\sin^4(\theta)+[1-\sin^2(\theta)]^2=\ldots$

That's brilliant. It so simple but I just couldn't see it, which is very frustrating. I'll work through that and see if I can do it.

Thank you.

**Furyan** · December 24th 2011, 10:06 AM

Originally Posted by Siron

Note that:
$\sin^4(\theta)+\cos^4(\theta)=\sin^4(\theta)+[1-\sin^2(\theta)]^2=\ldots$

That wasn't as easy as I thought it was going to be. My instinct was to write $\sin^4\theta = [1 - \cos^2\theta]^2$ as well, but that was a mess. I got there, following your work, by expanding and factoring the expression you gave me.

Thank you very much. There's so much to learn and you are all really helping a lot.

**Siron** · December 24th 2011, 10:11 AM

We have:
$\sin^4(\theta)+[1-\sin^2(\theta)]^2=\sin^4(\theta)+1-2\sin^2(\theta)+\sin^4(\theta)=1-2\sin^2(\theta)+2\sin^4(\theta)=1-2\sin^2(\theta)[1-\sin^2(\theta)]=1-2\sin^2(\theta)\cos^2(\theta)$

**Furyan** · December 24th 2011, 10:16 AM

Originally Posted by Siron

We have:
$\sin^4(\theta)+[1-\sin^2(\theta)]^2=\sin^4(\theta)+1-2\sin^2(\theta)+\sin^4(\theta)=1-2\sin^2(\theta)+2\sin^4(\theta)=1-2\sin^2(\theta)[1-\sin^2(\theta)]=1-2\sin^2(\theta)\cos^2(\theta)$

Thank you. Yes, that's how I got there in the end.

Only one to go, then integration with trig. functions, yipee!

**Siron** · December 24th 2011, 10:31 AM

You're welcome!

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